P\((A\cup B)\) = P(A) + P(B) - P(\(A\cap B\))
P\((A\cup B)\) = P(A) + P(B) - P(A)P(B) (since A and B are independent)
That is, 0.9 = 0.4 + P(B) − (0.4) P(B)
0.9 − 0.4 = (1− 0.4) P(B)
Therefore, P(B) = \(\frac{5}{6}\).

\(P(A)=\frac{3}{4}, P(B) =\frac{2}{5}, A \cup B=S \)
\(\Rightarrow P(A \cup B) =1\)
\(P(A \cap B) =P(A)+P(B)-P(A \cup B)\)
\(=\frac{3}{4}+\frac{2}{5}-1\)
\(=\frac{15+8}{20}-1=\frac{23-20}{20}=\frac{3}{20} \)
\(P(A / B) =\frac{P(A \cap B)}{P(B)} \)
\(=\frac{3 / 20}{2 / 5}=\frac{3}{8}\)

Given A and B are independent
\( \Rightarrow P(A \cap B) =P(A) P(B) \)
\(P(A \cup B) =0.6, P(A)=0.2\)
\(P(A \cup B) =P(A)+P(B)-P(A \cap B) \)
\(=P(A)+P(B)-P(A) \cdot P(B) \)
\(0 \cdot 6 =0 \cdot 2+P(B)(1-0 \cdot 2) \)
\(\frac{0 \cdot 4}{0 \cdot 8} =P(B) \)
\(\text { i.e., } P(B) =\frac{1}{2}=0.5\)

Given P(A) = 0.5, P(B) = 0.8
⇒ P(B/A) = 0.8
We kmow P(B/A) = \(\frac{P(A\cap B)}{P(A)}\)
⇒ 0.8 = \(\frac{P(A\cap B)}{0.5}\)
⇒ \(P(A\cap B)=(0.8)(0.5)=0.4\)
(i) Now P(A./B) = \(\frac{P(A\cap B)}{P(B)}=\frac{0.4}{0.8} =\frac{1}{2}=0.5\)
(ii) \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
= 0.5 + 0.8 - 0.4 = 1.3 - 0.4
= 0.9

\(\text {(i) } P(A)=0.4 \ \&\ P(A \cup B)=0.7\)
A and B are mutually exclusive then
\(P(A \cup B) =P(A)+P(B) \)
\(0.7 =0.4+P(B) \)
\(0.3 =P(B)\)
\(P(B) =0.3\)
(ii) A and B are independent then,
\(P(A \cup B) =P(A)+P(B)-P(A \cap B) \)
\(\therefore 0.7 =P(A)+P(B)-P(A) \cdot P(B) \)
\(0.7 =0.4+P(B)[1-0.4]\)
\(P(B) =\frac{0.7-0.4}{1-0.4}=\frac{0.3}{0.6}=\frac{1}{2}=0.5\)
\(\text { (iii) } P(A / B)=0.4\)
\(\frac{P(A \cap B)}{P(B)}=0.4\)
\(\Rightarrow \frac{P(A)+P(B)-P(A \cup B)}{P(B)}=0.4\)
\(\frac{0.4+P(B)-0.7}{P(B)}=0.4\)
\(P(B)-0.3=0.4 P(B)\)
\((1-0.4) P(B)=0.3\)
\(P(B)=\frac{0.3}{0.6}=\frac{1}{2}=0.5\)
\(\text {(iv) } P(B / A)=0.5\)
\(\frac{P(A \cap B)}{P(A)}=0.5\)
\(P(A)+P(B)-P(A \cup B)=0.5 P(A)\)
\(0.4+P(B)-0.7=0.5 \times 0.4\)
\(P(B)=0.2+0.3=0.5\)

Let B & G are the representation of boys & girls respectively.
Let F be the student taking first grade.
Then \(P(B)=\frac{2}{3}, P(G)=\frac{1}{3}\)
\(P(F / B)=0.70, P(F / G)=0.85\)
\(Now, P(F \cap B or F \cap G)=P(F \cap B)+P(F \cap G)\)
\(=P(B) \cdot P(F / B)+P(G) .P(F / G)\)
\(=\frac{2}{3} \times 0.7+\frac{1}{3} \times 0.85\)
\(=\frac{1.4+0.85}{3} \)
\( =\frac{2.25}{3}=\frac{225}{300} \)
\(=\frac{3}{4}=0.75\)

Let A be the oil change, P(A) = 0.3
B be the filter change, P(B) = 0.4
And \(P(A \cap B)=0.15\)
\(\text { (i) } P(B / A) =\frac{P(A \cap B)}{P(A)}\)
\(=\frac{0.15}{0.3}=\frac{15}{30}=0.5\)
\(\text { (ii) } P(A / B) =\frac{P(A \cap B)}{P(B)}\)
\(=\frac{0.15}{0.4}=\frac{15}{40}=\frac{3}{8}=0.375\)

Notice that three coins are tossed simultaneously = one coin is tossed three times.
The sample space S = { H,T } \(\times\) {H,T} \(\times\) {H,T}
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, n(S) = 8
Let A be the event of getting exactly one head, B be the event of getting atleast one head and C be the event of getting at most one head
A = {HTT, THT, TTH}; n(A) = 3
B = {HTT, THT, TTH, HHT, HTH, THH, HHH} n(B) = 7
C = {TTT, HTT, THT, TTH}; n(C) = 4
Therefore the required probabilities are
i) P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 8 } \)
ii) P(B) = \(\frac { n(B) }{ n(S) } =\frac { 7 }{ 8 } \)
iii)P(C) \(​​=\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}\)

Sample space S = {1, 2, 3, 4, 5, 6}.
Let A be the event of die shows an odd number.
Let B be the event of getting 5.
Then, A = {1, 3, 5}, B = {5}, and A\(\cap \)B = {5}.
Therefore, P(A) = \(\frac{3}{6}\) and P(A\(\cap \)B) = \(\frac{1}{6}\)
P(getting 5 / die shows an odd number) = P(B / A)
= \(\frac { P(A\cap B) }{ P(A) } =\frac { \frac { 1 }{ 6 } }{ \frac { 3 }{ 6 } } \)
P(B/A) = \(\frac{1}{3}\) .

The sample space is
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, n(S) = 10
Let A be the event of choosing an even number and B be the event of choosing an integer multiple of three.
A = {2, 4, 6, 8, 10}, n(A) = 5,
B = {3, 6, 9} , n(B) = 3
P(choosing an even integer) = P(A) = \(\frac { n(A) }{ n(S) } =\frac { 5 }{ 10 } =\frac { 1 }{ 2 } \)
P(choosing an integer multiple of three) = P(B) =\(\frac { n(B) }{ n(S) } =\frac { 3 }{ 10 } \)